equations of value - ut mathematics of value 1 single deposit under compound interest 2 investments...
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Equations of Value
1 Single Deposit Under Compound Interest
2 Investments with Multiple Contributions
Equations of Value
1 Single Deposit Under Compound Interest
2 Investments with Multiple Contributions
The amount function
• Let the principal be denoted by C . Then, in the usual notation, theamount function reads as
AC (T ) = C (1 + i)T for every T ≥ 0
• The above equation is referred to as the T time equation of value
• Knowing any three among the values AC (T ),C ,T , i allows us tocalculate the fourth one
• We will illustrate this fact through examples
The amount function
• Let the principal be denoted by C . Then, in the usual notation, theamount function reads as
AC (T ) = C (1 + i)T for every T ≥ 0
• The above equation is referred to as the T time equation of value
• Knowing any three among the values AC (T ),C ,T , i allows us tocalculate the fourth one
• We will illustrate this fact through examples
The amount function
• Let the principal be denoted by C . Then, in the usual notation, theamount function reads as
AC (T ) = C (1 + i)T for every T ≥ 0
• The above equation is referred to as the T time equation of value
• Knowing any three among the values AC (T ),C ,T , i allows us tocalculate the fourth one
• We will illustrate this fact through examples
The amount function
• Let the principal be denoted by C . Then, in the usual notation, theamount function reads as
AC (T ) = C (1 + i)T for every T ≥ 0
• The above equation is referred to as the T time equation of value
• Knowing any three among the values AC (T ),C ,T , i allows us tocalculate the fourth one
• We will illustrate this fact through examples
Unknown principal
• On July 1, 2012, you will need $1, 000. Assume that the annualeffective interest rate equals 0.0625. How much would you have todeposit on October 1, 2008 in order to be able to withdraw the$1, 000 from the account on July 1, 2012?
⇒ The start of the loan term is October 1, 2008. So, we call that timezero. Hence, the time at which we are given the accumulatedamount we want to produce is T = 3.75 (in years).
The equation of value reads as
1000 = C · (1 + 0.0625)3.75
Solving for C , we obtain C = 796.65.
Unknown principal
• On July 1, 2012, you will need $1, 000. Assume that the annualeffective interest rate equals 0.0625. How much would you have todeposit on October 1, 2008 in order to be able to withdraw the$1, 000 from the account on July 1, 2012?
⇒ The start of the loan term is October 1, 2008. So, we call that timezero. Hence, the time at which we are given the accumulatedamount we want to produce is T = 3.75 (in years).
The equation of value reads as
1000 = C · (1 + 0.0625)3.75
Solving for C , we obtain C = 796.65.
Unknown principal
• On July 1, 2012, you will need $1, 000. Assume that the annualeffective interest rate equals 0.0625. How much would you have todeposit on October 1, 2008 in order to be able to withdraw the$1, 000 from the account on July 1, 2012?
⇒ The start of the loan term is October 1, 2008. So, we call that timezero. Hence, the time at which we are given the accumulatedamount we want to produce is T = 3.75 (in years).
The equation of value reads as
1000 = C · (1 + 0.0625)3.75
Solving for C , we obtain C = 796.65.
Unknown principal
• On July 1, 2012, you will need $1, 000. Assume that the annualeffective interest rate equals 0.0625. How much would you have todeposit on October 1, 2008 in order to be able to withdraw the$1, 000 from the account on July 1, 2012?
⇒ The start of the loan term is October 1, 2008. So, we call that timezero. Hence, the time at which we are given the accumulatedamount we want to produce is T = 3.75 (in years).
The equation of value reads as
1000 = C · (1 + 0.0625)3.75
Solving for C , we obtain C = 796.65.
Unknown length of investment
• Find the length of time necessary for $1000 to accumulate to $1500if invested at 6% per annum compounded semiannually
⇒ It is more convenient to think of time-units in this situation ashalf-years. Let us denote the unknown number of half-years by n.
Then the equation of value reads as
1000(1 +0.06
2)n = 1500
We get n ≈ 13.7, i.e., the sought for length of time is about 6.85years.
Unknown length of investment
• Find the length of time necessary for $1000 to accumulate to $1500if invested at 6% per annum compounded semiannually
⇒ It is more convenient to think of time-units in this situation ashalf-years. Let us denote the unknown number of half-years by n.
Then the equation of value reads as
1000(1 +0.06
2)n = 1500
We get n ≈ 13.7, i.e., the sought for length of time is about 6.85years.
Unknown length of investment
• Find the length of time necessary for $1000 to accumulate to $1500if invested at 6% per annum compounded semiannually
⇒ It is more convenient to think of time-units in this situation ashalf-years. Let us denote the unknown number of half-years by n.
Then the equation of value reads as
1000(1 +0.06
2)n = 1500
We get n ≈ 13.7, i.e., the sought for length of time is about 6.85years.
Unknown length of investment
• Find the length of time necessary for $1000 to accumulate to $1500if invested at 6% per annum compounded semiannually
⇒ It is more convenient to think of time-units in this situation ashalf-years. Let us denote the unknown number of half-years by n.
Then the equation of value reads as
1000(1 +0.06
2)n = 1500
We get n ≈ 13.7, i.e., the sought for length of time is about 6.85years.
Unknown time: Doubling your money
• At a nominal interest rate of 10% convertible quarterly, how longwill it take you to double your money?
⇒ Because the wealth accumulated at any given time is proportional tothe principal, we can assume (without loss of generality) that theinitial investment is a single dollar.
Then the equation of value becomes
(1 +0.10
4)n = 2
where n denotes the unknown length of time (in quarter-years) thatit takes to double the initial investment
Solving for n, we get n ≈ 28.071 quarters of a year, i.e., the lengthof time we were looking for is about 7 years.
Unknown time: Doubling your money
• At a nominal interest rate of 10% convertible quarterly, how longwill it take you to double your money?
⇒ Because the wealth accumulated at any given time is proportional tothe principal, we can assume (without loss of generality) that theinitial investment is a single dollar.
Then the equation of value becomes
(1 +0.10
4)n = 2
where n denotes the unknown length of time (in quarter-years) thatit takes to double the initial investment
Solving for n, we get n ≈ 28.071 quarters of a year, i.e., the lengthof time we were looking for is about 7 years.
Unknown time: Doubling your money
• At a nominal interest rate of 10% convertible quarterly, how longwill it take you to double your money?
⇒ Because the wealth accumulated at any given time is proportional tothe principal, we can assume (without loss of generality) that theinitial investment is a single dollar.
Then the equation of value becomes
(1 +0.10
4)n = 2
where n denotes the unknown length of time (in quarter-years) thatit takes to double the initial investment
Solving for n, we get n ≈ 28.071 quarters of a year, i.e., the lengthof time we were looking for is about 7 years.
Unknown time: Doubling your money
• At a nominal interest rate of 10% convertible quarterly, how longwill it take you to double your money?
⇒ Because the wealth accumulated at any given time is proportional tothe principal, we can assume (without loss of generality) that theinitial investment is a single dollar.
Then the equation of value becomes
(1 +0.10
4)n = 2
where n denotes the unknown length of time (in quarter-years) thatit takes to double the initial investment
Solving for n, we get n ≈ 28.071 quarters of a year, i.e., the lengthof time we were looking for is about 7 years.
Unknown interest rate
• At what interest rate convertible quarterly would $1000 accumulateto $1600 in six years?
⇒ We need to find i (4) from the equation of value
1000(1 +i (4)
4)24 = 1600
We get
i (4)
4= (1.6)1/24 − 1 = 0.019776
and so i (4) = 0.0791.
• Assignment: Problems 2.2.1,2,3,6
Unknown interest rate
• At what interest rate convertible quarterly would $1000 accumulateto $1600 in six years?
⇒ We need to find i (4) from the equation of value
1000(1 +i (4)
4)24 = 1600
We get
i (4)
4= (1.6)1/24 − 1 = 0.019776
and so i (4) = 0.0791.
• Assignment: Problems 2.2.1,2,3,6
Unknown interest rate
• At what interest rate convertible quarterly would $1000 accumulateto $1600 in six years?
⇒ We need to find i (4) from the equation of value
1000(1 +i (4)
4)24 = 1600
We get
i (4)
4= (1.6)1/24 − 1 = 0.019776
and so i (4) = 0.0791.
• Assignment: Problems 2.2.1,2,3,6
Unknown interest rate
• At what interest rate convertible quarterly would $1000 accumulateto $1600 in six years?
⇒ We need to find i (4) from the equation of value
1000(1 +i (4)
4)24 = 1600
We get
i (4)
4= (1.6)1/24 − 1 = 0.019776
and so i (4) = 0.0791.
• Assignment: Problems 2.2.1,2,3,6
Unknown interest rate
• At what interest rate convertible quarterly would $1000 accumulateto $1600 in six years?
⇒ We need to find i (4) from the equation of value
1000(1 +i (4)
4)24 = 1600
We get
i (4)
4= (1.6)1/24 − 1 = 0.019776
and so i (4) = 0.0791.
• Assignment: Problems 2.2.1,2,3,6
Equations of Value
1 Single Deposit Under Compound Interest
2 Investments with Multiple Contributions
The time τ equation of value
• Consider a sequence of contributions to an investment account madeat times {tk} and represented by values {Ctk}
• This sequence of investments is liquidated at a time T and theresulting balance is denoted by B
• Suppose that a certain intermediate time τ bears a significance forthe investor and that the investor wants to value all thecontributions at that specific time τ
• Then the value of all the combined contributions to the accountvalued at time τ must be equal to the final balance B valued at timeτ
• Formally, we get the time τ equation of value∑k
Ctk
a(τ)
a(tk)= B
a(τ)
a(T )
The time τ equation of value
• Consider a sequence of contributions to an investment account madeat times {tk} and represented by values {Ctk}
• This sequence of investments is liquidated at a time T and theresulting balance is denoted by B
• Suppose that a certain intermediate time τ bears a significance forthe investor and that the investor wants to value all thecontributions at that specific time τ
• Then the value of all the combined contributions to the accountvalued at time τ must be equal to the final balance B valued at timeτ
• Formally, we get the time τ equation of value∑k
Ctk
a(τ)
a(tk)= B
a(τ)
a(T )
The time τ equation of value
• Consider a sequence of contributions to an investment account madeat times {tk} and represented by values {Ctk}
• This sequence of investments is liquidated at a time T and theresulting balance is denoted by B
• Suppose that a certain intermediate time τ bears a significance forthe investor and that the investor wants to value all thecontributions at that specific time τ
• Then the value of all the combined contributions to the accountvalued at time τ must be equal to the final balance B valued at timeτ
• Formally, we get the time τ equation of value∑k
Ctk
a(τ)
a(tk)= B
a(τ)
a(T )
The time τ equation of value
• Consider a sequence of contributions to an investment account madeat times {tk} and represented by values {Ctk}
• This sequence of investments is liquidated at a time T and theresulting balance is denoted by B
• Suppose that a certain intermediate time τ bears a significance forthe investor and that the investor wants to value all thecontributions at that specific time τ
• Then the value of all the combined contributions to the accountvalued at time τ must be equal to the final balance B valued at timeτ
• Formally, we get the time τ equation of value∑k
Ctk
a(τ)
a(tk)= B
a(τ)
a(T )
The time τ equation of value
• Consider a sequence of contributions to an investment account madeat times {tk} and represented by values {Ctk}
• This sequence of investments is liquidated at a time T and theresulting balance is denoted by B
• Suppose that a certain intermediate time τ bears a significance forthe investor and that the investor wants to value all thecontributions at that specific time τ
• Then the value of all the combined contributions to the accountvalued at time τ must be equal to the final balance B valued at timeτ
• Formally, we get the time τ equation of value∑k
Ctk
a(τ)
a(tk)= B
a(τ)
a(T )
The time 0 equation of value
• In the case that we choose τ = 0 the time 0 equation of valuebecomes ∑
k
Ctk v(tk) = B v(T )
• In words, it suffices to discount all the contributions and the balancefrom the times their corresponding cash flows took place to theinitial time 0
The time 0 equation of value
• In the case that we choose τ = 0 the time 0 equation of valuebecomes ∑
k
Ctk v(tk) = B v(T )
• In words, it suffices to discount all the contributions and the balancefrom the times their corresponding cash flows took place to theinitial time 0
The time T equation of value
• In the case we set τ = T , we get as the time T equation of value∑k
Ctk
a(T )
a(tk)= B
• That is, all the contributions must be augmented to their value atthe final time T
The time T equation of value
• In the case we set τ = T , we get as the time T equation of value∑k
Ctk
a(T )
a(tk)= B
• That is, all the contributions must be augmented to their value atthe final time T
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown payment
• In return for a promise to receive $600 at the end of 8 years, Rogeragrees to pay $100 at once, $200 at the end of 5 years, and to makea further payment X at the end of 10 years.
Assume that the nominal rate of interest is 8% convertiblesemiannually.
Find the payment at the end of 10 years.
⇒ Again, let us count time-periods in half-years.
Note that there is no particular liquidation moment; this means thatthere is no remaining balance after all the cash flows outlined in theproblem take place; we could choose any moment later than time 10and say that B = 0
Since we are looking for a payment that is made at time τ = 10, letus make that point in time our reference point. The equation ofvalue is
100(1.04)20 + 200(1.04)10 − 600(1.04)4 + X = 0
• We solve the above equation and get X = 186.76.
An example: Unknown rate of interest
• At what effective rate of interest will the present value of theinvestment consisting of $2000 deposited at the end of 2 years and$3000 deposited at the end of four years be equal to $4000?
⇒ The above wording implies that the reference point for ourcalculations should be time 0. Since this means that we arediscounting, it is more convenient to write the equation of value interms of the discount factor v .The time 0 equation of value is
2000v2 + 3000v4 = 4000
We get a quadratic in v2:
3v4 + 2v2 − 4 = 0
which yields only one positive solution v2 = 0.8685So,
(1 + i)2 = v−2 = 1.1514
and i = 0.073.
An example: Unknown rate of interest
• At what effective rate of interest will the present value of theinvestment consisting of $2000 deposited at the end of 2 years and$3000 deposited at the end of four years be equal to $4000?
⇒ The above wording implies that the reference point for ourcalculations should be time 0. Since this means that we arediscounting, it is more convenient to write the equation of value interms of the discount factor v .The time 0 equation of value is
2000v2 + 3000v4 = 4000
We get a quadratic in v2:
3v4 + 2v2 − 4 = 0
which yields only one positive solution v2 = 0.8685So,
(1 + i)2 = v−2 = 1.1514
and i = 0.073.
An example: Unknown rate of interest
• At what effective rate of interest will the present value of theinvestment consisting of $2000 deposited at the end of 2 years and$3000 deposited at the end of four years be equal to $4000?
⇒ The above wording implies that the reference point for ourcalculations should be time 0. Since this means that we arediscounting, it is more convenient to write the equation of value interms of the discount factor v .The time 0 equation of value is
2000v2 + 3000v4 = 4000
We get a quadratic in v2:
3v4 + 2v2 − 4 = 0
which yields only one positive solution v2 = 0.8685So,
(1 + i)2 = v−2 = 1.1514
and i = 0.073.
An example: Unknown rate of interest
• At what effective rate of interest will the present value of theinvestment consisting of $2000 deposited at the end of 2 years and$3000 deposited at the end of four years be equal to $4000?
⇒ The above wording implies that the reference point for ourcalculations should be time 0. Since this means that we arediscounting, it is more convenient to write the equation of value interms of the discount factor v .The time 0 equation of value is
2000v2 + 3000v4 = 4000
We get a quadratic in v2:
3v4 + 2v2 − 4 = 0
which yields only one positive solution v2 = 0.8685So,
(1 + i)2 = v−2 = 1.1514
and i = 0.073.
An example: Unknown rate of interest
• At what effective rate of interest will the present value of theinvestment consisting of $2000 deposited at the end of 2 years and$3000 deposited at the end of four years be equal to $4000?
⇒ The above wording implies that the reference point for ourcalculations should be time 0. Since this means that we arediscounting, it is more convenient to write the equation of value interms of the discount factor v .The time 0 equation of value is
2000v2 + 3000v4 = 4000
We get a quadratic in v2:
3v4 + 2v2 − 4 = 0
which yields only one positive solution v2 = 0.8685So,
(1 + i)2 = v−2 = 1.1514
and i = 0.073.
An example: Unknown payment times
• The present value of two payments of $100 each to be made at theend of n years and 2n years is $100. If i = 0.08, find n.
⇒ The equation of value (at time 0) is
100(1 + 0.08)n + 100(1 + 0.08)2n = 100,
i.e.,
(1 + 0.08)−n +((1 + 0.08)−n
)2= 1
Set A = (1.08)−n; the above equation becomes
A2 + A− 1 = 0
We only keep the positive solution A = 12 (√
5− 1)
and from there we get that n = − ln(A)/ ln(1.08) ≈ 6.25 years
An example: Unknown payment times
• The present value of two payments of $100 each to be made at theend of n years and 2n years is $100. If i = 0.08, find n.
⇒ The equation of value (at time 0) is
100(1 + 0.08)n + 100(1 + 0.08)2n = 100,
i.e.,
(1 + 0.08)−n +((1 + 0.08)−n
)2= 1
Set A = (1.08)−n; the above equation becomes
A2 + A− 1 = 0
We only keep the positive solution A = 12 (√
5− 1)
and from there we get that n = − ln(A)/ ln(1.08) ≈ 6.25 years
An example: Unknown payment times
• The present value of two payments of $100 each to be made at theend of n years and 2n years is $100. If i = 0.08, find n.
⇒ The equation of value (at time 0) is
100(1 + 0.08)n + 100(1 + 0.08)2n = 100,
i.e.,
(1 + 0.08)−n +((1 + 0.08)−n
)2= 1
Set A = (1.08)−n; the above equation becomes
A2 + A− 1 = 0
We only keep the positive solution A = 12 (√
5− 1)
and from there we get that n = − ln(A)/ ln(1.08) ≈ 6.25 years
An example: Unknown payment times
• The present value of two payments of $100 each to be made at theend of n years and 2n years is $100. If i = 0.08, find n.
⇒ The equation of value (at time 0) is
100(1 + 0.08)n + 100(1 + 0.08)2n = 100,
i.e.,
(1 + 0.08)−n +((1 + 0.08)−n
)2= 1
Set A = (1.08)−n; the above equation becomes
A2 + A− 1 = 0
We only keep the positive solution A = 12 (√
5− 1)
and from there we get that n = − ln(A)/ ln(1.08) ≈ 6.25 years
An example: Unknown payment times
• The present value of two payments of $100 each to be made at theend of n years and 2n years is $100. If i = 0.08, find n.
⇒ The equation of value (at time 0) is
100(1 + 0.08)n + 100(1 + 0.08)2n = 100,
i.e.,
(1 + 0.08)−n +((1 + 0.08)−n
)2= 1
Set A = (1.08)−n; the above equation becomes
A2 + A− 1 = 0
We only keep the positive solution A = 12 (√
5− 1)
and from there we get that n = − ln(A)/ ln(1.08) ≈ 6.25 years
An example: Unknown payment times
• The present value of two payments of $100 each to be made at theend of n years and 2n years is $100. If i = 0.08, find n.
⇒ The equation of value (at time 0) is
100(1 + 0.08)n + 100(1 + 0.08)2n = 100,
i.e.,
(1 + 0.08)−n +((1 + 0.08)−n
)2= 1
Set A = (1.08)−n; the above equation becomes
A2 + A− 1 = 0
We only keep the positive solution A = 12 (√
5− 1)
and from there we get that n = − ln(A)/ ln(1.08) ≈ 6.25 years
Method of Equated Time: Exact approach
• The type of problem we wish to consider next is the following:
An investor makes the contributions {Ctk ; k = 1, 2, . . . n} at times{tk}. Find the time T so that a single payment of C =
∑nk=1 Ctk at
time T has the same value at time 0 as the n contributions.
• First we formalize the conditions in the above problem:
CvT =n∑
k=1
Ctk vtk
• Hence,
vT =1
C
n∑k=1
Ctk vtk
and so
T =1
ln(v)· ln
(1
C
n∑k=1
Ctk vtk
)
Method of Equated Time: Exact approach
• The type of problem we wish to consider next is the following:
An investor makes the contributions {Ctk ; k = 1, 2, . . . n} at times{tk}. Find the time T so that a single payment of C =
∑nk=1 Ctk at
time T has the same value at time 0 as the n contributions.
• First we formalize the conditions in the above problem:
CvT =n∑
k=1
Ctk vtk
• Hence,
vT =1
C
n∑k=1
Ctk vtk
and so
T =1
ln(v)· ln
(1
C
n∑k=1
Ctk vtk
)
Method of Equated Time: Exact approach
• The type of problem we wish to consider next is the following:
An investor makes the contributions {Ctk ; k = 1, 2, . . . n} at times{tk}. Find the time T so that a single payment of C =
∑nk=1 Ctk at
time T has the same value at time 0 as the n contributions.
• First we formalize the conditions in the above problem:
CvT =n∑
k=1
Ctk vtk
• Hence,
vT =1
C
n∑k=1
Ctk vtk
and so
T =1
ln(v)· ln
(1
C
n∑k=1
Ctk vtk
)
Method of Equated Time: Exact approach
• The type of problem we wish to consider next is the following:
An investor makes the contributions {Ctk ; k = 1, 2, . . . n} at times{tk}. Find the time T so that a single payment of C =
∑nk=1 Ctk at
time T has the same value at time 0 as the n contributions.
• First we formalize the conditions in the above problem:
CvT =n∑
k=1
Ctk vtk
• Hence,
vT =1
C
n∑k=1
Ctk vtk
and so
T =1
ln(v)· ln
(1
C
n∑k=1
Ctk vtk
)
Method of Equated Time: Theapproximation
• An approximation to the above solution is
T̄ =1
C
n∑k=1
Ctk tk =n∑
k=1
Ctk
Ctk
• The above weighted average of payment times is called the methodis equated time approximation
• It is worth noticing that we consistently have
T̄ ≥ T
Method of Equated Time: Theapproximation
• An approximation to the above solution is
T̄ =1
C
n∑k=1
Ctk tk =n∑
k=1
Ctk
Ctk
• The above weighted average of payment times is called the methodis equated time approximation
• It is worth noticing that we consistently have
T̄ ≥ T
Method of Equated Time: Theapproximation
• An approximation to the above solution is
T̄ =1
C
n∑k=1
Ctk tk =n∑
k=1
Ctk
Ctk
• The above weighted average of payment times is called the methodis equated time approximation
• It is worth noticing that we consistently have
T̄ ≥ T
An Example
• Payments of $100, $200 and $500 are due at the ends of years 2, 3and 8, respectively. Assuming an effective interest rate of 5%, findthe time t∗ at which a single payment of $800 would be equivalentto the above combination of payments.
⇒ Using the method of equated time, we get an approximation to thetime t∗, i.e.,
t̄∗ =100
800· 2 +
200
800· 3 +
500
800· 8 = 6 years
Our exact model yields the following sequence of calculations:The present values of the three payments are
100 · 0.90703, 200 · 0.86384, 500 · 0.67684
So,
1
800(100 · 0.90703 + 200 · 0.86384 + 500 · 0.67684) = 0.75236
Finally,
t∗ = − ln(0.75236)
ln(1.05)= 5.83 years
An Example
• Payments of $100, $200 and $500 are due at the ends of years 2, 3and 8, respectively. Assuming an effective interest rate of 5%, findthe time t∗ at which a single payment of $800 would be equivalentto the above combination of payments.
⇒ Using the method of equated time, we get an approximation to thetime t∗, i.e.,
t̄∗ =100
800· 2 +
200
800· 3 +
500
800· 8 = 6 years
Our exact model yields the following sequence of calculations:The present values of the three payments are
100 · 0.90703, 200 · 0.86384, 500 · 0.67684
So,
1
800(100 · 0.90703 + 200 · 0.86384 + 500 · 0.67684) = 0.75236
Finally,
t∗ = − ln(0.75236)
ln(1.05)= 5.83 years
An Example
• Payments of $100, $200 and $500 are due at the ends of years 2, 3and 8, respectively. Assuming an effective interest rate of 5%, findthe time t∗ at which a single payment of $800 would be equivalentto the above combination of payments.
⇒ Using the method of equated time, we get an approximation to thetime t∗, i.e.,
t̄∗ =100
800· 2 +
200
800· 3 +
500
800· 8 = 6 years
Our exact model yields the following sequence of calculations:The present values of the three payments are
100 · 0.90703, 200 · 0.86384, 500 · 0.67684
So,
1
800(100 · 0.90703 + 200 · 0.86384 + 500 · 0.67684) = 0.75236
Finally,
t∗ = − ln(0.75236)
ln(1.05)= 5.83 years
An Example
• Payments of $100, $200 and $500 are due at the ends of years 2, 3and 8, respectively. Assuming an effective interest rate of 5%, findthe time t∗ at which a single payment of $800 would be equivalentto the above combination of payments.
⇒ Using the method of equated time, we get an approximation to thetime t∗, i.e.,
t̄∗ =100
800· 2 +
200
800· 3 +
500
800· 8 = 6 years
Our exact model yields the following sequence of calculations:The present values of the three payments are
100 · 0.90703, 200 · 0.86384, 500 · 0.67684
So,
1
800(100 · 0.90703 + 200 · 0.86384 + 500 · 0.67684) = 0.75236
Finally,
t∗ = − ln(0.75236)
ln(1.05)= 5.83 years
An Example
• Payments of $100, $200 and $500 are due at the ends of years 2, 3and 8, respectively. Assuming an effective interest rate of 5%, findthe time t∗ at which a single payment of $800 would be equivalentto the above combination of payments.
⇒ Using the method of equated time, we get an approximation to thetime t∗, i.e.,
t̄∗ =100
800· 2 +
200
800· 3 +
500
800· 8 = 6 years
Our exact model yields the following sequence of calculations:The present values of the three payments are
100 · 0.90703, 200 · 0.86384, 500 · 0.67684
So,
1
800(100 · 0.90703 + 200 · 0.86384 + 500 · 0.67684) = 0.75236
Finally,
t∗ = − ln(0.75236)
ln(1.05)= 5.83 years
An Example
• Payments of $100, $200 and $500 are due at the ends of years 2, 3and 8, respectively. Assuming an effective interest rate of 5%, findthe time t∗ at which a single payment of $800 would be equivalentto the above combination of payments.
⇒ Using the method of equated time, we get an approximation to thetime t∗, i.e.,
t̄∗ =100
800· 2 +
200
800· 3 +
500
800· 8 = 6 years
Our exact model yields the following sequence of calculations:The present values of the three payments are
100 · 0.90703, 200 · 0.86384, 500 · 0.67684
So,
1
800(100 · 0.90703 + 200 · 0.86384 + 500 · 0.67684) = 0.75236
Finally,
t∗ = − ln(0.75236)
ln(1.05)= 5.83 years
Assignments
• Assignment: See Example 2.3.6 for a problem which is not to easyto tackle without a calculatorProblems 2.3.4,5, 9, 10